JEE Main 2021 (Online) 27th July Morning Shift | Capacitor Question 68 | Physics

A capacitor of capacitance C = 1 $$\mu$$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $$\Omega$$. The time taken for the capacitor to be charged to get 50 V is :

[Take ln 2 = 0.69]

1.44 $$\times$$ 10^$$-$$4 s

3.33 $$\times$$ 10^$$-$$4 s

0.69 $$\times$$ 10^$$-$$4 s

0.30 $$\times$$ 10^$$-$$4 s

JEE Main 2021 (Online) 27th July Morning Shift

MCQ (Single Correct Answer)

-1

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be :

(Given area of plate = A)

JEE Main 2021 (Online) 27th July Morning Shift Physics - Capacitor Question 70 English

JEE Main 2021 (Online) 27th July Morning Shift Physics - Capacitor Question 70 English

$${{15} \over {34}}{{K{\varepsilon _0}A} \over d}$$

$${{15} \over 6}{{K{\varepsilon _0}A} \over d}$$

$${{25} \over 6}{{K{\varepsilon _0}A} \over d}$$

$${9 \over 6}{{K{\varepsilon _0}A} \over d}$$

JEE Main 2021 (Online) 27th July Morning Shift

MCQ (Single Correct Answer)

-1

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

$${V \over {K + 2}}$$

$${V \over K}$$

$${{3V} \over {K + 2}}$$

$${{3V} \over K}$$

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

If q_f is the free charge on the capacitor plates and q_b is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q_b an be expressed as :

$${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$$

$${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$$

$${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$$

$${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$$

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